## Example №1. Solving of a System of Linear Equations by the Gauss Elimination (One Solution)This solution has been made using the calculator presented on the site. Example №2. Solving of a system of linear equations by the Gauss elimination (many solutions) Example №3. Solving of a system of linear equations by the Gauss elimination (no solution) Example №4. Solving of a system of linear equations by the Gauss Jordan elimination (one solution) Example №5. Solving of a system of linear equations by the Gauss Jordan elimination (many solutions) Please note that the coefficients will disappear which located in the "red" positions.
The equation 1 multiplied by -1 is added to the equation 2. more info ( 5 x _{1} + 4 x_{1} * ( -1) ) + ( 3 x _{2} + 2 x_{2} * ( -1) ) + ( -5 x _{3} + ( -3 x_{3}) * ( -1) ) = -8 + ( -3) * ( -1) This transformation will allow us to count without fractions for some time.
The equation 2 and equation 1 are reversed.
The equation 1 multiplied by -4 is added to the equation 2. more info ( 4 x _{1} + x_{1} * ( -4) ) + ( 2 x _{2} + x_{2} * ( -4) ) + ( -3 x _{3} + ( -2 x_{3}) * ( -4) ) = -3 + ( -5) * ( -4) The "red" coefficient is zero.
The equation 1 multiplied by -4 is added to the equation 3. more info ( 4 x _{1} + x_{1} * ( -4) ) + ( x _{2} + x_{2} * ( -4) ) + ( 5 x _{3} + ( -2 x_{3}) * ( -4) ) = 22 + ( -5) * ( -4) The "red" coefficient is zero.
The equation 3 multiplied by -1 is added to the equation 2. more info ( -2 x _{2} + ( -3 x_{2}) * ( -1) ) + ( 5 x _{3} + 13 x_{3} * ( -1) ) = 17 + 42 * ( -1) This transformation will allow us to count without fractions for some time.
The equation 2 multiplied by 3 is added to the equation 3. more info ( -3 x _{2} + x_{2} * 3 ) + ( 13 x _{3} + ( -8 x_{3}) * 3 ) = 42 + ( -25) * 3 The "red" coefficient is zero.
We will find the variable x _{3} from equation 3 of the system. - 11 x _{3} = - 33 x _{3} = 3 We will find the variable x _{2} from equation 2 of the system. x _{2} - 8 x_{3} = - 25 x _{2} = - 25 + 8 x_{3} x _{2} = - 25 + 8 * ( 3 )x _{2} = - 1 We will find the variable x _{1} from equation 1 of the system. x _{1} + x_{2} - 2 x_{3} = - 5 x _{1} = - 5 - x_{2} + 2 x_{3} x _{1} = - 5 - ( - 1 ) + 2 * ( 3 )x _{1} = 2 Result: x _{1} = 2 x _{2} = - 1 x
_{3} = 3 © 2010-2021 If you have any comments, please write to siteReshmat@yandex.ru |